Question

The value of $\int_{0}^{\frac{\pi}{2}}$ log($\frac{4+3\,sin\,x}{4+3\,cos\,x}$)dx is

• A. 2
• B. $\frac{3}{4}$
• C. 0
• D. -2

Answer 1

Answer: (C)

0

Answer 2

Let I=$\int_{0}^{\frac{\pi}{2}}$ log($\frac{4+3\,sin\,x}{4+3\,cos\,x}$)dx………………....(1)

⇒I=$\int_{0}^{\frac{\pi}{2}}$log[$\frac{4+3\,sin\,(\frac{\pi}{2}-x)}{4+3\,cos\,(\frac{\pi}{2}-x)}$]dx ($\int_{0}^{a}$ƒ(x)dx=$\int_{0}^{a}$ƒ(a-x)dx)

⇒I=$\int_{0}^{\frac{\pi}{2}}$ log($\frac{4+3\,cos\,x}{4+3\,sin\,x}$)dx…………………...(2)

Adding(1)and(2),we obtain

2I=$\int_{0}^{\frac{\pi}{2}}${log($\frac{4+3\,sin\,x}{4+3\,cos\,x}$)+log($\frac{4+3\,cos\,x}{4+3\,sin\,x}$)}dx

⇒2I=$\int_{0}^{\frac{\pi}{2}}$ log1dx

⇒2I=$\int_{0}^{\frac{\pi}{2}}$0dx

⇒I=0

Hence, the correct Answer is C.